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Use suffix notation, summing repeated indices, and the Levi-Civita symbol convention $(A\times B{)}_i={ϵ}_{ijk}{A}_j{B}_k$. Contracting the outer symbol with the symbol from the second cross product gives the epsilon-delta identitywhere $\delta$ is the Kronecker delta. ThusThe second term is zero because $A_p{A}_{\ell }$ is symmetric in $p,\ell$ and the Levi-Civita symbol is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the scalar triple product $A\cdot (B\times C)$. HenceThis cross products with a common vector identity requires no independence assumption, so it also covers zero or parallel vectors.

Three vectors in three-dimensional Euclidean space are non-coplanar precisely when their scalar triple product is nonzero. Put $\Delta =a\cdot (b\times c)\ne 0$. In particular, $a\ne 0$. The identity for cross products with a common vector givesFor completeness, this follows from the vector triple product identity: writing $p=a\times b$ gives $p\times (a\times c)=a(p\cdot c)-c(p\cdot a)$, with $p\cdot c=\Delta$ and $p\cdot a=0$. Therefore the new scalar triple product isThus $a,a\times b,a\times c$ are non-coplanar, or equivalently linearly independent.

By the vector triple product identity,The last inner product vanishes by perpendicularity. Taking the inner product with $d$ givesThis is the identity for cross products with a common vector, and holds without nondegeneracy assumptions.